Calculus 04 Transcendental Functions 2up

7/30/2019 Calculus 04 Transcendental Functions 2up

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4.7 Derivatives of the exponential and logarithmic functions 77

Thus whether x is positive or negative, the derivative is the same.

What about the functions ax and loga x? We know that the derivative of ax is some

constant times ax itself, but what constant? Remember that “the logarithm is the expo-

nent” and you will see that a = eln a. Then

ax = (elna)x = ex lna,

and we can compute the derivative using the chain rule:

d

dxa

x

=

d

dx (e

lna

)

x

=

d

dxe

x lna

= (ln a)e

x lna

= (lna)a

x

.

The constant is simply ln a. Likewise we can compute the derivative of the logarithm

function loga x. Since

x = elnx,

we can take the logarithm base a of both sides to get

loga(x) = loga(elnx) = lnx loga e.

Then

ddx loga x = 1x loga e.

Finally, since

a = elna

loga(a) = loga(elna) = lna loga e

1 = lna loga e

1

lna= loga e,

we can replace loga e to get

ddx loga x = 1x lna.

You may if you wish memorize the formulas

d

dxax = (lna)ax and

d

dxloga x =

1

x lna.

Because the “trick” a = eln a is often useful, and sometimes essential, it may be better to

remember the trick, not the formula.

78 Chapter 4 Transcendental Functions

EXAMPLE 4.5 Compute the derivative of f (x) = 2x.

d

dx2x =

d

dx(eln 2)x

=d

dxex ln2

=

d

dxx ln 2

ex ln 2

= (ln2)ex ln 2

EXAMPLE 4.6 Compute the derivative of f (x) = 2x2

= 2(x2).

d

dx2x

2

=d

dxex

2 ln 2

=

d

dxx2 ln 2

ex

2 ln 2

= (2ln2)xex2 ln 2

= (2ln2)x2x2

EXAMPLE 4.7 Compute the derivative of f (x) = xx. At first this appears to be a

new kind of function: it is not a constant power of x, and it does not seem to be an

exponential function, since the base is not constant. But in fact it is no harder than the

previous example.d

dxxx =

d

dxex lnx

= d

dx

x lnx ex lnx

= (x1

x+ lnx)xx

= (1 + lnx)xx

EXAMPLE 4.8 Recall that we have not justified the power rule except when the

exponent is a positive or negative integer. We can use the exponential function to take





4.8 Limits revisited 81

EXAMPLE 4.12 Compute limx→π

x2 − π2

sinxin two ways.

First we use L’Hopital’s Rule: Since the numerator and denominator both approach

zero,

limx→π

x2 − π2

sinx= lim

x→π

2x

cosx,

provided the latter exists. But in fact this is an easy limit, since the denominator now

approaches −1, so

limx→π

x2 − π2

sinx=

2π

−1

= −2π.

We don’t really need L’Hopital’s Rule to do this limit. Rewrite it as

limx→π

(x + π)x− π

sinx

and note that

limx→π

x− π

sinx= lim

x→π

x− π

− sin(x− π)= lim

x→0− x

sinx

since x− π approaches zero as x approaches π. Now

limx→π(x + π)x−π

sinx = limx→π(x + π) limx→0−x

sinx = 2π(−1) = −2π

as before.

EXAMPLE 4.13 Compute limx→∞

2x2 − 3x + 7

x2 + 47x + 1in two ways.

As x goes to infinity both the numerator and denominator go to infinity, so we may

apply L’Hopital’s Rule:

limx→∞

2x2 − 3x + 7

x2

+ 47x + 1

= limx→∞

4x− 3

2x + 47

.

In the second quotient, it is still the case that the numerator and denominator both go to

infinity, so we are allowed to use L’Hopital’s Rule again:

limx→∞

4x− 3

2x + 47= lim

x→∞

4

2= 2.

So the original limit is 2 as well.


Again, we don’t really need L’Hopital’s Rule, and in fact a more elementary approach

is easier—we divide the numerator and denominator by x2:

limx→∞

2x2 − 3x + 7

x2 + 47x + 1= lim

x→∞

2x2 − 3x + 7

x2 + 47x + 1

1x2

1x2

= limx→∞

2− 3x + 7

x2

1 + 47x + 1

x2.

Now as x approaches infinity, all the quotients with some power of x in the denominator

approach zero, leaving 2 in the numerator and 1 in the denominator, so the limit again is

2.

EXAMPLE 4.14 Compute limx→0

secx− 1sinx

.

Both the numerator and denominator approach zero, so applying L’Hopital’s Rule:

limx→0

secx− 1

sinx= lim

x→0

secx tanx

cosx=

1 · 0

1= 0.

EXAMPLE 4.15 Compute limx→0+

x lnx.

This doesn’t appear to be suitable for L’Hopital’s Rule, but it also is not “obvious”.

As x approaches zero, lnx goes to

−∞, so the product looks like (something very small)

·(something very large and negative). But this could be anything: it depends on how small

and how large . For example, consider (x2)(1/x), (x)(1/x), and (x)(1/x2). As x approaches

zero, each of these is (something very small) · (something very large), yet the limits are

respectively zero, 1, and ∞.

We can in fact turn this into a L’Hopital’s Rule problem:

x lnx =lnx

1/x=

lnx

x−1.

Now as x approaches zero, both the numerator and denominator approach infinity (one

−∞ and one +∞, but only the size is important). Using L’Hopital’s Rule:

limx→0+

lnx

x−1= lim

x→0+

1/x

−x−2 = limx→0+

1

x(−x2) = lim

x→0+−x = 0.

One way to interpret this is that since limx→0+

x lnx = 0, the x approaches zero much faster

than the lnx approaches −∞.





4.9 Implicit Differentiation 85

We might have recognized at the start that (1,−√ 3) is on the function y = L(x) =

−√ 4− x2. We could then take the derivative of L(x), using the power rule and the chain

rule, to get

L′(x) = −1

2(4 − x2)−1/2(−2x) =

x√ 4− x2

.

Then we could compute L′(1) = 1/√

3 by substituting x = 1.

Alternately, we could realize that the point is on L(x), but use the fact that y′ = −x/y.

Since the point is on L(x) we can replace y by L(x) to get

y′

= −x

L(x) = −x

√ 4− x2 ,

without computing the derivative of L(x) explicitly. Then we substitute x = 1 and get the

same answer as before.

In the case of the circle it is possible to find the functions U (x) and L(x) explicitly, but

there are potential advantages to using implicit differentiation anyway. In some cases it is

more difficult or impossible to find an explicit formula for y and implicit differentiation is

the only way to find the derivative.

EXAMPLE 4.17 Find the derivative of any function defined implicitly by yx2

+e

y

= x.We treat y as an unspecified function and use the chain rule:

d

dx(yx2 + ey) =

d

dxx

(y · 2x + y′ · x2) + y′ey = 1

y′x2 + y′ey = 1 − 2xy

y′(x2 + ey) = 1− 2xy

y′ =1− 2xy

x2 + ey

You might think that the step in which we solve for y′ could sometimes be difficult—

after all, we’re using implicit differentiation here because we can’t solve the equation

yx2 + ey = x for y, so maybe after taking the derivative we get something that is hard to

solve for y′. In fact, this never happens. All occurrences y′ come from applying the chain

rule, and whenever the chain rule is used it deposits a single y′ multiplied by some other

expression. So it will always be possible to group the terms containing y′ together and


factor out the y′, just as in the previous example. If you ever get anything more difficult

you have made a mistake and should fix it before trying to continue.

It is sometimes the case that a situation leads naturally to an equation that defines a

function implicitly.

EXAMPLE 4.18 Consider all the points (x, y) that have the property that the distance

from (x, y) to (x1, y1) plus the distance from (x, y) to (x2, y2) is 2a (a is some constant).

These points form an ellipse, which like a circle is not a function but can viewed as two

functions pasted together. Because we know how to write down the distance between two

points, we can write down an implicit equation for the ellipse:

(x− x1)2 + (y − y1)2 +

(x− x2)2 + (y − y2)2 = 2a.

Then we can use implicit differentiation to find the slope of the ellipse at any point.

EXAMPLE 4.19 We have already justified the power rule by using the exponential

function, but we could also do it for rational exponents by using implicit differentiation.

Suppose that y = xm/n, where m and n are positive integers. We can write this implicitly

as yn = xm, then b ecause we justified the power rule for integers, we can take the derivative

of each side:

nyn−1y′ = mxm−1

y′ =m

n

xm−1

yn−1

y′ =m

n

xm−1

(xm/n)n−1

y′ =m

nxm−1−(m/n)(n−1)

y′ =m

nxm−1−m+(m/n)

y′ =m

n

x(m/n)−1





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Calculus 04 Transcendental Functions 2up